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Rudik [331]
4 years ago
10

A 100-kg object is moving at 20 m/s when a force brings the object to rest 3 points

Physics
2 answers:
densk [106]4 years ago
6 0
0.5 i just took the test
zhenek [66]4 years ago
4 0

Answer:

0.5

Explanation:

By reducing the force, the time taken to stop the object increases, as they are inversely proportional.

Therefore, when a doubling of the time is observed, a halving of the force should be as well.

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Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5​
juin [17]

Answer:

The ratio of lengths of the two mathematical pendulums is 9:4.

Explanation:

It is given that,

The ratio of periods of two pendulums is 1.5

Let the lengths be L₁ and L₂.

The time period of a simple pendulum is given by :

T=2\pi \sqrt{\dfrac{l}{g}}

or

T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}

Where

l is length of the pendulum

l\propto T^2

or

\dfrac{l_1}{l_2}=(\dfrac{T_1}{T_2})^2 ....(1)

ATQ,

\dfrac{T_1}{T_2}=1.5

Put in equation (1)

\dfrac{l_1}{l_2}=(1.5)^2\\\\=\dfrac{9}{4}

So, the ratio of lengths of the two mathematical pendulums is 9:4.

3 0
3 years ago
What is magnification
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The action or process of magnifying something or being magnified, especially visually. Hope this helped
5 0
4 years ago
Find the vector sum of the 3 vectors. A= -3i - 2j + 7k, B= i + 3j + 3k, B= -5k​
DENIUS [597]

Answer:

34k+B plus 9 :)

Explanation:

7 0
3 years ago
Which branch deals with anything foreign (other countries)<br> A executive b legislative c judicial
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6 0
3 years ago
Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one en
Tom [10]

Explanation:

Wavelength in an emission spectrum,  \lambda=435\ nm=435\times 10^{-9}\ m  

The energy of an electron is given by :

E=\dfrac{hc}{\lambda}

Where

h is the Planck's constant

c is the speed of light

For 435 nm, the energy of the electron will be :

E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8\ m/s}{435\times 10^{-9}}

E=4.57\times 10^{-19}\ J

We know that 1\ eV=1.6\times 10^{-19}\ J

So, E=\dfrac{4.57\times 10^{-19}}{1.6\times 10^{-19}}

So, E = 2.86 eV

The energy of the electron dropping from one energy level is 2.86 eV. We know that,

\dfrac{hc}{\lambda}=E_{n_2}-E_{n_1}

From the given energy levels :

E_5-E_2=-0.544-(-3.403)

E_5-E_2=2.859\ eV

So, the transition must be from E₅ to E₂. Hence, this is the required solution.                                             

5 0
4 years ago
Read 2 more answers
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